Solve x^5+9x^2+bx+9=(x-5)^2(x-2)

$$x ^ { 5 } + 9 x ^ { 2 } + b x + 9 = ( x - 5 ) ^ { 2 } ( x - 2 )$$

Solve for b

$$b=-x^{4}+x^{2}-21x+45-\frac{59}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{5}+9x^{2}+bx+9=\left(x^{2}-10x+25\right)\left(x-2\right)$$

Use the distributive property to multiply x^{2}-10x+25 by x-2 and combine like terms.

$$x^{5}+9x^{2}+bx+9=x^{3}-12x^{2}+45x-50$$

Subtract x^{5} from both sides.

$$9x^{2}+bx+9=x^{3}-12x^{2}+45x-50-x^{5}$$

Subtract 9x^{2} from both sides.

$$bx+9=x^{3}-12x^{2}+45x-50-x^{5}-9x^{2}$$

Combine -12x^{2} and -9x^{2} to get -21x^{2}.

$$bx+9=x^{3}-21x^{2}+45x-50-x^{5}$$

Subtract 9 from both sides.

$$bx=x^{3}-21x^{2}+45x-50-x^{5}-9$$

Subtract 9 from -50 to get -59.

$$bx=x^{3}-21x^{2}+45x-59-x^{5}$$

The equation is in standard form.

$$xb=-x^{5}+x^{3}-21x^{2}+45x-59$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-x^{5}+x^{3}-21x^{2}+45x-59}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-x^{5}+x^{3}-21x^{2}+45x-59}{x}$$

Divide x^{3}-21x^{2}+45x-59-x^{5} by x.

$$b=-x^{4}+x^{2}-21x+45-\frac{59}{x}$$